Problem: Find the point in the $xz$-plane that is equidistant from the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1).$
Solution: Since the point lies in the $xz$-plane, it is of the form $(x,0,z).$  We want this point to be equidistant to the points $(1,-1,0),$ $(2,1,2),$ and $(3,2,-1),$ which gives us the equations
\begin{align*}
(x - 1)^2 + 1^2 + z^2 &= (x - 2)^2 + 1^2 + (z - 2)^2, \\
(x - 1)^2 + 1^2 + z^2 &= (x - 3)^2 + 2^2 + (z + 1)^2.
\end{align*}These equations simplify to $2x + 4z = 7$ and $4x - 2z = 12.$  Solving these equation, we find $x = \frac{31}{10}$ and $z = \frac{1}{5},$ so the point we seek is $\boxed{\left( \frac{31}{10}, 0, \frac{1}{5} \right)}.$